CODEWITHSUNDEEP
c++templatesc++11
Alex Brooks
Alex Brooks

Reputation: 1151

Extracting return type from function template type

I have a wrapper function similar to the following:

template<typename Func, typename ... Args>
void func(Func f, Args ... args) {
  ...
  f(args...);
  ...
}

Is it possible to extract the return type from the type Func at compile time?

For example, a user has the code and calls func in the following manner:

template<typename T>
T add(const T& l, const T& r) {
  return l + r;
}

int main(int argc, char** argv) {
  double a = 4.5, b = 5.5;
  func(add<double>, a, b);
  return 0;
}

Can we infer in the invocation of func that a function was passed to it which returns a double? I would like to use the return type information for other things inside func.

Upvotes: 1

Views: 2663

Answers (3)

Chronial
Chronial

Reputation: 70843

Since c++17 it's std::invoke_result:

using result_type = typename std::invoke_result_t<Func(Args...)>;

Upvotes: 2

dau_sama
dau_sama

Reputation: 4367

If you have access to C++14, or just out of curiosity, try not to commit to a particular type. The following code uses a generic lambda and the type is deduced for you. In many cases I find it much easier to work with.

#include <iostream>
using namespace std;

template<typename Func, typename ... Args>
void func(Func&& f, Args ... args) {
  return;
}

auto add = [](auto l, auto r) { return l * r; };

int main() 
{
    auto a = double{4.5};
    auto b = double{5.5};
    func(add, a, b);
    return 0;
}

Ideone: http://ideone.com/wXnc0g

Upvotes: 0

Praetorian
Praetorian

Reputation: 109279

Either of the following should work

using result_type = decltype(std::declval<Func>()(std::declval<Args>()...));

or

using result_type = typename std::result_of<Func(Args...)>::type;

Live demo

Upvotes: 6

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